Examples

Example
A mixture of 100 k mole which contain 60% mole percent n-pentane and 40% n-heptane are vaporized at 101.32 kpa pressure under different conditions unitl 60 k mole is distilled .
what is the average composition of total vapour distillated (Xd) and composition of the liquid (Xw), the equilibrium is given as:
xA 1 0.86 0.59 0.398 0.25 0.195 0.059
yA 1 0.98 0.925 0.856 0.701 0.521 0.271
Repeat (a), if the equilibrium data is given by the yA =1.8 xA
Repeat (a), if the relative volatility ?AB = 5.8


Solution
Let A n-pentane , F =100 k mole , Xf =60%= 0.6 , D = 60 k mole
W=F-D W = 100-60 =40 k mole
xA yA 1 / (xA - yA)
0.6 0.92 3.125
0.5 0.89 2.56
0.4 0.83 2.32
0.3 0.75 2.22
0.25 0.7 2.22
0.2 0.63 2.32
0.1 0.42 3.12

Area under the curve = 18 * 0.1 * 0.5 = 0.9
ln w/f=ln 100/40=0.916=?_xf^xw?dxA/(yA-xA)
From the curve Xw = 0.22
DXd =FXf – W Xw , Xd = 100(0.6) – 40(0.22) / 60 = 0.853
If the eq. data is given y = mx+C , yA =1.8 xA
ln F / W = -1 / m-1 ln[ (m-1) xm +c / (m-1) xf +c ] , mx 1.8 , c=0

ln 100/ 40= -1 / 0.8 ln 0.8 Xw / 0.8 Xf , Xw =0.288
Xd= FXf-WXw / D = 100(0.6)-40(0.288) / 60 =0.808
If the ?AB is given ?AB = 5.8
ln??w/f?= (-1)/(?AB-1) ln?[Xw(1-Xf)/Xf(1-Xw) ]+ln??[ (1-xf)/(1-xw)?]
-0.916=1 / 5.8-1 ln [ xw(1-0.6) / 0.6(1-xw)]+ ln [ (1-0.6) / (1-xw)]
Xw = 0.26 by try and error , assume the Xw less than Xf
Continuous / Flash distillation (Equilibrium distillation) :

F = V + L , F Xf = V yA + L XA , where
F = mole per unit time of feed of mole fraction Xf of A
V = mole per unit time of vapour formed with yA , the mole fraction of A
L = mole per unit time of liquid with XA, the mole fraction of A
Example / Amixture of 100 kmole which contain 60% mole percent n-pentaine and 40% n-heptane is vaporized at 101.32 kpa pressure in an equilibrium or / flash distillate and 40 kmole are distillated , what is the composition of the vapour distillated yA and of the liquid left XA, U.L.E is as in ?
XA 1` 0.86 0.59 0.398 0.25 0.145 0.059
yA 1 0.98 0.925 0.836 0.701 0.521 0.271
Solution
Flash equilibrium distillation single stage distillation . F= 100 kmole , D= 40 kmole

F = W+D , F Xf = V yA + L XA , then 100(0.6)= 60(Xw)+ 40 (Xd)
Let Xw= 0.43 less than 0.6 Xf
100(0.6)= 60(0.43) + 40(Xd) , then Xd = 0.855 ? 0.86 in the table
60 = 25.8+ 34.2 because Xd ? 0.86 and Xw= 0.43
Continuous distillation
Rectification can be classified in to two parts
Binary ( two components) 2- Multi components
The operation of a typical fractionating column may be followed by reference to Figure 9. The column consists of a cylindrical structure divided into sections by a series of perforated trays which permit the upward flow of vapour. The liquid reflux flows across each tray, over a weir and down a downcomer to the tray below. The vapour rising from the top tray passes to a condenser and then through an accumulator or reflux drum and a reflux divider, where part is withdrawn as the overhead product D, and the remainder is returned to the top tray as reflux R.

Fig .9 Continuous fractionating column with rectifying and stripping sections
Number of plates required in a distillation column :
There are two basic methods for determining the number of plates required. The first is due to SORELand later modified by LEWIS, and the second is due to MCCABE and THIELE. The Lewis method is used here for binary systems. This method is also the basis of modern computerised methods. The McCabe–Thiele method is particularly . important since it introduces the idea of the operating line which is an important common concept in multistage operations.
LEWIS –SOREL ( equations)
McCabe–Thiele method ( charts)
McCabe–Thiele method ( charts)
Top operating line


At top , the materials balance the above number of plats (n): Vn = L n+1 + D
yn Vn= Xn+1 L n+1 Xd D then yn=( Ln+1 / Vn ) Xn+1 +( D / Vn)Xd , since the molar liquid over flow is constant, then ( Ln = Ln+1) , therefore yn=( Ln / Vn ) Xn+1 +( D / Vn) Xd , if Xn+1 = Xd Then yn=( Ln / Vn ) Xd +( D / Vn) Xd
first point (Xd , Xd) , second point ( 0, Xd / Rt ) or (0, D Xd / Vn)